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umka2103 [35]
3 years ago
8

Complete the analogy solid:Table:.......water​

Chemistry
1 answer:
prisoha [69]3 years ago
4 0

Answer:

of

Explanation:

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An increase in the surface area of reactants in a heterogeneous reaction will result in what?
marusya05 [52]
It will result in an increase in the rate of rxn
4 0
2 years ago
Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

[Pb^{2+}]=3.9 \times 10^{-2}M

this is the concentration required to initiate precipitation

Explanation:

PbCl_2  ⇄ Pb^{2+}+2Cl^-

Precipitation starts when ionic product is greater than solubility product.

Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

Ksp=2.4\times 10^{-4}

lets solubility=S

[Pb^{2+}] = S

[Cl^-]=2S

Ksp=[Pb^{2+}]\times [Cl^-]^2

Ksp=S \times (2S)^2

Ksp=4S^3

S=\sqrt[3]{\frac{Ksp}{4} }

S=3.9\times 10^{-2}

[Pb^{2+}]=3.9 \times 10^{-2}M this is the concentration required to initiate precipitation

4 0
3 years ago
We can also use the equation for enthalpy change for physical phase changes. Consider the phase change H2O(l) → H2O(g). Calculat
BlackZzzverrR [31]
<span>44.0 kJ is the answer</span>
4 0
3 years ago
Read 2 more answers
"Write the electron Configuration expected for element 113 and the configurations for the two cations it is most likely to form"
arlik [135]

Answer:

Element:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Cations:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

Explanation:

The electron configuration is the distribution of the electrons in the sublevels in order of the crescent energy of them. The crescent energy of the sublevels follows the Linus Pauling's diagram, which is attached below. The sublevel "s" comports until 2 electrons, the sublevel "p" until 6 electrons, the sublevel "d" until 10 electrons, and sublevel "f" until 14 electrons.

So, for the element with an atomic number of 113, the neutral atom will have 113 electrons:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹

Thus the element is at the 7 period (the highest level), and group 13 (most energic sublevel p with 1 electron), the group of the aluminum. It needs to lose 3 electrons to be stable and follow the octet rule, but the subshells of the last shell are too far away in energetic order, thus, it most probably to lose the electron of 7p and form a monovalent cation, and can lose the two electrons of 7s to form a trivalent cation:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰

5 0
3 years ago
What is the volume of 70.0 g of ether if the density of ether is 0.70 g/mL?
frosja888 [35]
M=70.0 g
p=0.70 g/mL

v=m/p

v=70.0/0.70=100.00 mL
8 0
2 years ago
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